Handout #7 1 Homework 2 - Solutions 1 Problem 3-7 (Dally and Poulton) Buses Without Stubs: One can build a bus (multidrop transmission line) in which there are no reflections off the stubs by placing matching networks at each drop across the line, as shown in Figure 3-57. (a)Design the resistive matching network, N, shown in the figure so that a signal transmitted to the network from any of its three terminals is propagated out the other two terminals (possibly attenuated) with no reflections. (b) With the impedance values shown in the figure, 20-Ω bus and 100-Ω stubs, how much energy is lost from the signal traveling down the bus at each stub? (c) How much energy would be lost if both the stubs and the bus were 50-Ω lines? 100Ω 20Ω N 20Ω Figure 1: Multidrop Bus (a) We can make a network of 3 resistors 100Ω R1 20Ω 20Ω R2 R2 Thus, we must solve for the following equivalences: 20Ω = R2 + [(R2 + 20Ω)==R1 + 100Ω)] 100Ω = R1 + [(R2 + 20Ω)==(R2 + 20Ω)] Solving these equations simultaneously, we get: R1 = 89:5Ω Handout #7 2 R2 = 1:05Ω (b) To solve for the energy lost at each of stubs, we solve for the current and voltage left after 1 stub. 100Ω V original C Ioriginal R1 20Ω 20Ω R2 A R2 B VA VB IA IB We start by solving for the current and voltage at point A, shown in the figure above. The voltage at point A will be: Z1 VA = × Voriginal = 0:947 × Voriginal R2 + Z1 ; whereZ1 = (R2 + 20Ω)==(R1 + 100Ω) = 18:9Ω The current flowing from point A to point B will be: R1 + 100Ω IA = × Ioriginal = 0:9 × Ioriginal R1 + 100Ω + R2 + 20Ω The voltage and current at point B will be: 20Ω VB = × Voriginal = 0:95 × VA = 0:89Voriginal 20Ω + R2 IB = IA Thus, the total power left to pass onto the next stub on the bus is: Pout = IB ∗ VB = 0:81 ∗ Poriginal So, the power lost at each stub as the signal travels down the line is: 19%energy loss at each stub We note that 5% of Poriginal continues onto the 100Ω stub at point C, so the actual energy loss we have through our network is only 14% . However, the energy loss to the signal continuing down the rest of the bus is, as above stated 81% of Poriginal. (c) Now, if the stubs and the bus were all 50 Ohm, we can go through the exact same procedure to find the resistor values, which turn out to be: R1 = 16:7Ω R2 = 16:7Ω Handout #7 3 We note that since the network is now symmetric in three ways, we would expect that R1 = R2 as indeed our results show. Again, using the same procedure as above, we get: 75%energy loss at each stub Or, in other words. Only 25% of the original energy, Poriginal, is available for the proceeding stubs. Again, as noted above, 25% of the original power also reaches point C in our circuit. So, only 50% of our energy is consumed in our resistive network. However, only 25% of our original signal energy is available to continue on the bus to the rest of the stubs. We can see the trade-off between the first and second networks. If there are many stubs on the bus, we want the maximum signal power available after passing each stub, as in the first setup. This is especially true if the transmitter stub is far away from the receiver stub. However, if there are few stubs, we want the maximum power to reach the stubs, as in the second setup. Since a normal bus has a lot of stubs attached to it, the first setup of unequal impedances between stub and bus segment is generally a better idea. Handout #7 4 2 Problem 3-11 (Dally and Poulton) Frequency-dependent Termination: (1) impedance vs frequency 1 1 (a) Z = R + Cj! = 50 + 10nF j! (b) Z = j!L + R = j!10nH + 50 (c) Z = j!L + R = j!10nH + 50 From the equations above, we can plot the graph. (b) and (c) have the same impedance(therefore has the same plot). (a) 150 100 Zout (lin) (a) |Z| 200k 400k 600k 800k 1x Frequency (lin) (HERTZ) (a) 50 (a) Re(Z) 0 -50 (a) Im(Z) Zout (lin) -100 -150 200k 400k 600k 800k 1x Frequency (lin) (HERTZ) (b) & (c) 80 70 (b), (c) |Z| Zout (lin) 60 50 0 200x 400x 600x 800x 1g Frequency (lin) (HERTZ) (b) & (c) 60 (b), (c) Re(Z) 40 Zout (lin) 20 (b), (c) Im(Z) 0 0 200x 400x 600x 800x 1g Frequency (lin) (HERTZ) Figure 2: Impedance vs Frequency (2) waveform received and reflected (a) At the discontinuity, capacitor act like an short circuit and becomes an open circuit in steady state. At time = 0, there's no reflection since terminal act like matched impedance. But the reflection coefficient goes to 1 in steady state as the capacitor acts like open. Here τ = RC = (50 + 50) ∗ 10nF = 1µs, it means that it takes more than 1µs for the output voltage charged up to final state. Since there's no voltage peaking by the reflection, there's not a big difference between risetime = 100ps and risetime = 1ns. Handout #7 5 * file: p311-2a.sp * file: p311-2a.sp 1.8 1.8 1.6 1.6 1.4 1.4 1.2 1.2 received received 1 1 800m 800m Voltages (lin) Voltages (lin) 600m 600m 400m 400m 200m 200m 0 0 0 500n 1u 1.5u 2u 0 500n 1u 1.5u 2u Time (lin) (TIME) Time (lin) (TIME) * file: p311-2a.sp * file: p311-2a.sp 900m 900m 800m 800m 700m 700m 600m 600m 500m 500m Reflected Reflected 400m 400m Result (lin) Result (lin) 300m 300m 200m 200m 100m 100m 0 0 0 500n 1u 1.5u 2u 0 500n 1u 1.5u 2u Time (lin) (TIME) Time (lin) (TIME) (a) rise time = 100ps (b) rise time = 1ns Figure 3: Problem 3-11 (a) (b) Since the inductor acts a an open circuit, the reflected voltage shoots up initially, but eventually goes back down to 0 (when the inductor starts acting as a short circuit). Summing of incident wave and reflection wave is divided to resistor and inductor, and received signal is only on the resistor part. Actual received waveform is R Z − Z0 R 2Z (1 + t ) = ( t ) j!L + R Zt + Z0 j!L + R Zt + Z0 So, it start at 0 when incident wave arrives(at discontinuity point V(! = 1)=0 ), and goes to 1 at L 10n steady state. Time constant for this curve is τ = R = 50+50 = 0:1ns The amplitude of incident wave is 1V(amplitude on the left side of the termination, that is, 2V at the source when it has 50Ohm input resistance with 50Ohm transmission line) with rise time 100ps and 1ns. (b) received (b) received 1 1 800m 800m 600m 600m Voltages (lin) 400m Voltages (lin) 400m 200m 200m 0 0 0 1n 2n 3n 4n 5n 0 1n 2n 3n 4n 5n Time (lin) (TIME) Time (lin) (TIME) (b)reflected (b) reflected 650m 100m 600m 550m 500m 80m 450m 400m 60m 350m 300m Result (lin) Result (lin) 250m 40m 200m 150m 20m 100m 50m 0 0 0 1n 2n 3n 4n 5n 0 1n 2n 3n 4n 5n Time (lin) (TIME) Time (lin) (TIME) (a) rise time = 100ps (b) rise time = 1ns Figure 4: Problem 3-11 (b) The overshoot of reflected wave is depend on the rise time of the incident wave. So there is a big difference between these two waveforms. Handout #7 6 (c) In the beginning, when the inductor acts as an open circuit, it does not let all the voltage through, hence the voltage at the right is less than one. Eventually, the inductor becomes a short circuit and the voltage goes to 1. The peak overshoot can be calculated as ∆V τ t = ( )[1 − exp(− r )] = 0:63 V tr τ When rise time = 100ps , ∆V = 0:63, when rise time = 1ns, ∆V = 0:0095. (c) received (c) received 1.6 1 1.4 1.2 800m 1 600m 800m Voltages (lin) Voltages (lin) 600m 400m 400m 200m 200m 0 0 0 1n 2n 3n 4n 5n 0 1n 2n 3n 4n 5n Time (lin) (TIME) Time (lin) (TIME) (c) reflected (c) reflected 650m 100m 600m 550m 500m 80m 450m 400m 60m 350m 300m Result (lin) Result (lin) 250m 40m 200m 150m 20m 100m 50m 0 0 0 1n 2n 3n 4n 5n 0 1n 2n 3n 4n 5n Time (lin) (TIME) Time (lin) (TIME) (a) rise time = 100ps (b) rise time = 1ns Figure 5: Problem 3-11 (c) 3 Problem 3-16 (Dally and Poulton) Extracting Parasitics: Develop a model circuit composed of ideal transmission lines, inductors, and capacitors that gives the same response as the given waveform. You will want to simulate your model circuit with HSPICE to verify correspondence. There isn't a clean analytical way of generating the given waveform. We can start with a few equations that give rough estimates of the device values, but because of secondary effects, intermediate reflections, oscillations between inductors and capacitors, etc., eventually we have to resort to the brute force method of trying various parameters until we get a satisfying circuit. The first step is to determine the arrangement of devices that will exhibit the same behavior as expressed by the waveform. Figure 3-47 of the textbook shows the reflection behavior of a large capacitor and a large inductor.